14 marksGanita Prakash Cl-8 Part 2, Section 6.2, pages 135-136
A trick says: think of a number, double it, add four, divide by two, subtract the number you first thought of — and the answer is always 2. Explain with algebra why this works. How would you change the steps so the answer is always 3? Always 5? Can you build a longer trick that still ends at a fixed value?
Hint. Carry the unknown through every step as a letter instead of trying particular numbers.
Why the trick works. Do not test numbers one at a time — carry a letter through the steps and the answer appears once and for all.
| Step | Expression |
|---|
| Think of a number | x |
| Double it | 2x |
| Add four | 2x + 4 |
| Divide by two | x + 2 |
| Subtract the original number | (x + 2) − x = 2 |
The x cancels at the last step, so whatever you started with, the answer is 2. Try x = 7: 14 → 18 → 9 → 2 ✓. Try x = 100: 200 → 204 → 102 → 2 ✓. Try x = 0.5: 1 → 5 → 2.5 → 2 ✓.
Where the 2 comes from. The 4 that was added gets halved to 2, and nothing else survives. So the final answer is always half of the number you added.
To make the answer 3: add 6 instead of 4. Check: x → 2x → 2x + 6 → x + 3 → 3 ✓.
To make the answer 5: add 10. Check: x → 2x → 2x + 10 → x + 5 → 5 ✓.
In general, to finish at k, add 2k.
A longer trick that still ends at a fixed value. Any sequence of steps whose net effect on x is zero will do. For instance:
| Step | Expression |
|---|
| Think of a number | x |
| Add 5 | x + 5 |
| Multiply by 3 | 3x + 15 |
| Subtract 9 | 3x + 6 |
| Divide by 3 | x + 2 |
| Subtract the original number | 2 |
| Multiply by 4 | 8 |
The design rule is simple: make sure the coefficient of x is 1 just before you subtract the original number. Everything after that is arithmetic on a constant, so it can be made as elaborate as you like without breaking the trick.
✦ The trick works because the algebra reduces to (x + 2) − x = 2, so the starting number cancels. Adding 6 instead of 4 makes the answer 3, adding 10 makes it 5, and adding 2k makes it k. Longer tricks work the same way: arrange the steps so the coefficient of x is 1 when you subtract the original number.
24 marksGanita Prakash Cl-8 Part 2, Section 6.2, pages 136-137
Mukta thinks of a date. She multiplies the month by 5, adds 6, multiplies by 4, adds 9, multiplies by 5, then adds the day, and reports 291. How did Shubham work out that the date was 26/01? Explain the algebra behind the trick.
Hint. Track the month as M and the day as D through every step, and see what number the steps build.
Follow the steps with letters. Let the month be M and the day be D.
| Step | Expression |
|---|
| Multiply the month by 5 | 5M |
| Add 6 | 5M + 6 |
| Multiply by 4 | 20M + 24 |
| Add 9 | 20M + 33 |
| Multiply by 5 | 100M + 165 |
| Add the day | 100M + 165 + D |
Why this shape is exactly what the trick needs. The month has been multiplied up to 100M, and the day is added on top as D. Since a day is at most 31, it occupies only the last two digits, while the month sits in the hundreds and above. Strip off the constant 165 and the month and day fall apart cleanly.
Decoding 291.
- 291 = 100M + 165 + D
- 291 − 165 = 126, so 100M + D = 126
- D ≤ 31 takes the last two digits: D = 26; what remains is M = 1
So the date was the 26th of January — Republic Day.
Check it forwards. M = 1, D = 26: 5 → 11 → 44 → 53 → 265 → 265 + 26 = 291 ✓
A second example from the chapter. Mukta reports 1390. Then 1390 − 165 = 1225, so M = 12 and D = 25 — the 25th of December.
Why the constant is 165. Trace the added numbers: the 6 gets multiplied by 4 and then by 5, giving 120; the 9 gets multiplied by 5, giving 45; and 120 + 45 = 165. Change either added number and the constant to subtract changes with it.
✦ The steps build the number 100M + 165 + D. Shubham subtracted 165 from 291 to get 126; since a day needs at most two digits, the last two digits give D = 26 and what remains gives M = 1 — the 26th of January.
34 marksGanita Prakash Cl-8 Part 2, Section 6.2, page 137
Using the same date trick, find the dates whose final answers are (i) 1269, (ii) 394, (iii) 296. Can the steps of the trick be changed and the date still be recovered?
Hint. Subtract the constant first, then split what is left into a month part and a two-digit day part.
The rule. The final answer is 100M + 165 + D, so subtract 165 and read off the last two digits as the day and the rest as the month.
(i) 1269. 1269 − 165 = 1104 → last two digits 04, what remains 11 → 4th of November.
Check: M = 11, D = 4: 55 → 61 → 244 → 253 → 1265 → 1265 + 4 = 1269 ✓
(ii) 394. 394 − 165 = 229 → last two digits 29, what remains 2 → 29th of February.
Check: M = 2, D = 29: 10 → 16 → 64 → 73 → 365 → 365 + 29 = 394 ✓
Worth noticing: 29 February exists only in a leap year, so this is a date that occurs once every four years.
(iii) 296. 296 − 165 = 131 → last two digits 31, what remains 1 → 31st of January.
Check: M = 1, D = 31: 5 → 11 → 44 → 53 → 265 → 265 + 31 = 296 ✓
Can the steps be changed? Yes — and this is the useful part. Two conditions must hold:
- The month must end up multiplied by 100 (or more). Here the multipliers are 5 × 4 × 5 = 100, so the month is pushed clear of the two digits the day needs.
- You must know the constant to subtract, which is whatever the added numbers grow into.
For example, replace 'add 6' by 'add 3' and 'add 9' by 'add 7'. Then the expression becomes 5M → 5M + 3 → 20M + 12 → 20M + 19 → 100M + 95 → 100M + 95 + D. So you would subtract 95 instead of 165, and everything else works identically.
You could even use different multipliers, provided they still make 100: multiply by 10, then by 5, then by 2, for instance. The trick survives as long as the month lands in the hundreds and you keep track of the constant.
✦ (i) 4 November (1269 − 165 = 1104) (ii) 29 February (394 − 165 = 229) (iii) 31 January (296 − 165 = 131). The steps can be changed freely so long as the month is finally multiplied by 100 and you subtract whatever constant the added numbers grow into — for example, adding 3 and 7 instead of 6 and 9 means subtracting 95 instead of 165.
Solutions written by the tuition.in editorial team and checked against NCERT Ganita Prakash Grade 8 Part 2 (hegp206.pdf), Chapter 6 'Algebra Play', pages 135-147. HAND-WRITTEN throughout. Part 2 books carry NO printed answer key, so every answer was derived from first principles and independently recomputed in Python. FIGURES READ OFF HIGH-DPI RENDERS AND RE-SOLVED: the three p.138 pyramids (bottom rows 6,2 / 3,4,3 / 5,4,5,0 giving tops 8, 14 and 32); the p.138 top-down pyramid 10 / 4,6 / 1,3,3; the p.139 letter-number pyramid 60 / 32,28 / 12,20,8; the three p.139 four-row pyramids, whose bottom rows solve to 4,9,6,1 (top 50), 5,14,7,2 (top 70) and 3,5,5,2 (top 35); the p.140 Figure it Out bottom rows 4,13,8 / 7,11,3 / 10,14,25 (tops 38, 32, 63) and 8,19,21,13 / 7,18,19,6 / 9,7,5,11 (tops 141, 124, 56); and the two p.142 algebra grids, which solve to square 11 and circle 5 with a third-row total of 21, and circle 4 and diamond 7 with a third-row total of 15 (the fourth row of that grid is printed blank for the student to design). The Virahanka-Fibonacci result was proved and then checked by computation for n = 2 to 6: every entry of the pyramid is a Virahanka-Fibonacci number, row k from the bottom is the consecutive run V(2k-1) to V(n+k-1), and the top is V(2n-1) - so a 29-row pyramid tops out at V57 = 591286729879. All six arrangements were enumerated for each Largest Product question: 32 x 5 = 160, 31 x 7 = 217 and 53 x 9 = 477, each beating its runner-up by exactly the margin the algebra predicts (4, 4 and 12). The date trick decodes as 100M + 165 + D, giving 4 November, 29 February and 31 January for 1269, 394 and 296. The puzzle answers were each verified by substitution: 7 flowers with 8 per shrine, 20 horses and 35 hens, a daughter of 6 and mother of 30, Gauri 6 cows and Naina 12, a dosa price of Rs 80 or a target of 175 dosas, the fraction sequence equal to 1/3 because n^2/(4n^2 - n^2) = 1/3, and Karim starting with 7 coins with the genie's general ruinous charge being c = 2^k x / (2^k - 1).. Questions are referenced from the NCERT textbook for identification.
